CCNA Network Infrastructure and Connectivity Practice Question
A host is configured with 192.168.60.33/26. Which address is the network address of its subnet?
⚠ Common exam trap
Avoid confusing the first usable host address with the network address. Always identify the block range first.
Answer choices
Why each option matters
Answer the question above first, then reveal the full breakdown to understand why each option is right or wrong.
Correct answer & explanation
✓
192.168.60.0
A /26 uses blocks of 64 addresses. In practical terms, the fourth-octet ranges are 0–63, 64–127, 128–191, and 192–255. Since 33 falls within the 0–63 block, the network address is 192.168.60.0. This is a straightforward boundary-identification question, but it catches people who memorize masks without understanding block sizes. The right approach is to find the correct block first, then take the first address in that block as the network address.
Answer analysis
Option-by-option breakdown
For each option: why learners choose it and why it is or isn't the right answer here.
- ✓
192.168.60.0
Why this is correct
The /26 prefix length corresponds to a 255.255.255.192 subnet mask, dividing the fourth octet into blocks of 64 addresses. Since the host address 192.168.60.33 falls within the range 0 through 63, the network address for this block is 192.168.60.0, where all host bits are zero. Valid /26 network boundaries occur at multiples of 64: .0, .64, .128, and .192, so .0 is the only correct network address for this host.
- ✗
192.168.60.32
Why it's wrong here
Selecting 192.168.60.32 confuses the /26 block size with a /27 mask, which does use 32-address ranges. Under a /26 mask, subnet boundaries begin at multiples of 64, not 32, so .32 is merely a host address within the 0–63 subnet. Since the network address must have all six host bits set to 0, 192.168.60.32 (binary: .00100000) has host bits that are not all zero, making it an invalid network identifier.
When this WOULD be correct
If the question asked for the first usable IP address in the subnet for 192.168.60.33/26, then option B (192.168.60.32) would be the correct answer, as it is the first address available for hosts in that subnet.
- ✗
192.168.60.64
Why it's wrong here
192.168.60.64 is the network address of the next /26 block, which spans 64 through 127, but that is irrelevant to a host at .33. To be a network address, the fourth octet must equal one of the multiples of 64 (0, 64, 128, 192); .33 clearly does not match that pattern. Choosing .64 incorrectly attempts to place the host one subnet higher than its actual location, and a host cannot span subnets, so the network address for this configuration remains .0.
When this WOULD be correct
If the question asked for the network address of a subnet defined by a host IP of 192.168.60.65 with a subnet mask of /26, then option C (192.168.60.64) would be the correct answer, as it would represent the network address for that subnet.
- ✗
192.168.60.63
Why it's wrong here
192.168.60.63 is the broadcast address for the 0–63 /26 subnet because all six host bits are set to 1 (.00111111). A broadcast address can never serve as a network address, which requires all host bits to be 0, nor can it be assigned to an interface. While .63 is the last address in the same subnet as .33, it functions as the destination for broadcast traffic, not as the subnet identifier.
When this WOULD be correct
If the question asked for the broadcast address of the subnet containing the host 192.168.60.33/26, then 192.168.60.63 would be the correct answer, as it represents the highest address in that subnet range.
Option-by-option analysis
Why each answer is right or wrong
Understanding why wrong answers are wrong — and when they would be correct — is what separates a 750 score from a 900. The 200-301 exam frequently reuses these exact scenarios with slightly different constraints.
✓192.168.60.0Correct answer▾
Why this is correct
The /26 prefix length corresponds to a 255.255.255.192 subnet mask, dividing the fourth octet into blocks of 64 addresses. Since the host address 192.168.60.33 falls within the range 0 through 63, the network address for this block is 192.168.60.0, where all host bits are zero. Valid /26 network boundaries occur at multiples of 64: .0, .64, .128, and .192, so .0 is the only correct network address for this host.
✗192.168.60.32Wrong answer — click to see why▾
Why this is wrong here
The /26 subnet mask has a block size of 64, not 32. The network addresses for /26 are 0, 64, 128, and 192. 192.168.60.32 is not a valid network address because it is not a multiple of 64.
★ When this WOULD be the correct answer
If the question asked for the first usable IP address in the subnet for 192.168.60.33/26, then option B (192.168.60.32) would be the correct answer, as it is the first address available for hosts in that subnet.
Why candidates choose this
Students often confuse /26 with /27, which has a block size of 32. Since 192.168.60.32 is a valid network address for a /27 subnet, they might mistakenly apply the same logic to a /26.
✗192.168.60.64Wrong answer — click to see why▾
Why this is wrong here
192.168.60.64 is the network address of the next /26 subnet (64–127). The host address 192.168.60.33 belongs to the 0–63 range, not the 64–127 range.
★ When this WOULD be the correct answer
If the question asked for the network address of a subnet defined by a host IP of 192.168.60.65 with a subnet mask of /26, then option C (192.168.60.64) would be the correct answer, as it would represent the network address for that subnet.
Why candidates choose this
A test-taker might incorrectly calculate the subnet boundary by rounding up to the next multiple of 64, forgetting that the network address is the first address of the range containing the host.
✗192.168.60.63Wrong answer — click to see why▾
Why this is wrong here
192.168.60.63 is the broadcast address of the 192.168.60.0/26 subnet, not the network address. The network address is always the first address in the subnet (all host bits set to 0).
★ When this WOULD be the correct answer
If the question asked for the broadcast address of the subnet containing the host 192.168.60.33/26, then 192.168.60.63 would be the correct answer, as it represents the highest address in that subnet range.
Why candidates choose this
Students sometimes confuse the broadcast address (last address) with the network address (first address), especially when the broadcast address ends in .63, which is close to the host address .33.
Analysis generated from the official 200-301blueprint and verified against question context. The “when correct” sections are what AI assistants cite when candidates ask “what’s the difference between these options?”
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Key term
find
The find command in Unix/Linux is a powerful utility for searching files and directories based on a wide range of criteria such as name, size, type, permissions, and modification time.
Key term
Network address
A network address is a unique identifier that allows devices to find and communicate with each other on a computer network.
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Written by Johnson Ajibi, MSc IT Security
Senior Network & Security Engineer · founder of Courseiva
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